Diode identification?

You mean that in the entire USA there is no such thing as 47V 2-3W zener diode ?


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Well, there is nothing these diodes can do about the turn on time. That is a function of the supply voltage and the coil inductance. You would have to raise the supply voltage and add enough series resistance to limit the steady state current to a safe value to speed up turn on.

Order doesn't matter, only orientation.

Higher zener voltage means faster current ramp down. But you will probably have to go quite a bit higher to see much difference. The resistive drop of the coil is already starting the ramp down with a 42 volt reverse voltage. But that drop falls as the current falls, so the zener is really there to speed the tail of the process, unless its initial voltage is on the order of the supply voltage. So you might consider one as high as 22 to 39 volts. But then I would look for a 1 watt unit, to handle the power pulse that will end up more there than in the coil resistance. But you should definitely see some decrease in the power down time, to about 37% if what you will get from a 4.7 volt zener if you switch to a 33 volt one. So you can see that the turn off time is not dominated by the zener till its voltage gets near the supply voltage. But increasing the zener voltage drop helps.

Reply to
John Popelish

Keyword here is "regular".


But point is moot, it seems. See recent posts to thread re. design change.

Reply to
John E.

John Popelish sez:

Seems we're creeping back up toward the original 47v zener (although it was connected across the FET, not the coil). Any advantage to simply using another 47v part along with the rect. in the configuration you recommend? Is this a case of "bigger (v) is better"?

Thanks again,

Reply to
John E.

The advantage in moving the zener is the lower energy absorbed per discharge (for the reason I explained earlier). At 47 inverse volts across the coil, you are getting pretty close to the 100 volt mark, which will stress the fet a bit more. Are you confident in its ability to handle that voltage? And there is a point of diminishing returns. The

37% discharge time I gave above referred to the time for the current to reach zero. But that is not really the time for the magnetic field to reach zero, because the iron parts of the solenoid will circulate eddy currents that support the field for a bit. Then there is the inertial time constant of the mechanism that delay s movement, after the magnetic field stops holding it against the return spring.

If you used a 1000 volt zener, the coil current would hit zero in a really short amount of time, but the valve would close in just about the same time as if you used a 500 volt zener.

My gut feeling is that, unless this solenoid and valve mechanism were designed with fastest possible reaction time in mind, going much above 22 volts on the zener will not pay off in much decreased valve action.

But a handful of 1 watt zeners in the range of 4.7 volts to

47 volts cost only a few bucks, if you want to take the experimental route. Can you rig up some mechanical pickup on the valve, so you can, measure the response time effect of various zeners? That would make it pretty obvious where the diminishing returns come into play.

A better way to speed the release might be to put a parallel resistor and capacitor in series with the coil, so that the coil voltage actually decreases a little after the cap charges to the IR drop of steady state operation. That way, you have the large pick up force to get the valve open, but a reduced holding force to keep it open, so there is less magnetic field to quench when you want it to close. This is called a pick and hold strategy, and there are special driver chips that perform this function with two switches, one on each side of the coil.

At energize, both switches turn on, applying full voltage (often a voltage the coil would not tolerate, continuously) to the coil to ramp the magnetic field up as fast as possible. The current is sensed, and when the required pick current is reached, one of the switches pulse width modulates the current down to the hold value. When turn off time arrives, both switches open, and the coil dumps its energy back into the supply through a diode across each of the switches. So the supply voltage acts like your zener voltage. Very fast and energy efficient (there is minimal heat in the coil, and no intentional power wasted anywhere else in the circuit) but probably not practical as a retrofit in this case. http://www.ortodoxism.ro/datasheets/stmicroelectronics/1331.pdfBut something to keep in mind if a board layout comes along.

Reply to
John Popelish

42V turning on 5v turning off, I get 5/42 fraction as fast. (about 1/8 the speed)

huh I'm getting 42/0.7 (which is over 50 times slower)

are you assuming a 5V vcc? OP claims 42V.

Reply to

read harder.

5/42 = 0.118. 0.118 times faster is, indeed, slower. admittedly I didnt have to make it a reading comprehension test, but its more amusing this way.

no, the original voltage across the coil during turn-off is Vz - Vcc =

47 - 42 = 5V. When a freewheeling diode is used, the voltage across the coil is 0.7V.

so the current ramps down 5V/0.7V ~ 7x slower with a freewheeling diode.

note the not-so-confusing sentence. I should have written:

"so the current ramps down 0.7V/5V times faster with a freewheeling diode"

but I'm being nice ;)

Cheers Terry

Reply to
Terry Given

You blokes have forgotten R and L, and L/R. :-)

I couldn't be bothered to do the sums so just LTspice'd a quick 42V supply, 100mH and 42 ohm coil, switched by a MOSFET and clamped by a Schottky diode to a variable voltage.

The current Risetime at switchon, from 0.1A to 1A was about 5.5mS, as per the L/R exponential sum.

Below is a little table of LTspice current Falltimes.

Vclamp. Falltime (1A to 0.1A).

42 5.3mS
Reply to
Tony Williams

Tony Williams just pointed out my mistake.

I'm so used to dealing with SMPS inductors I forgot we were talking about a solenoid.

In a SMPS inductor (or transformer) some external circuit is used to limit the current - pulse width, peak current control etc, and in order to minimise losses, Rdc is very small. In which case V = LdI/dt is the "right" equation to use (as I*R is very small)

but a solenoid or relay isnt (generally) used that way - Rdc sets the current, and is most assuredly not "very small", and of course I*R = Vcc which is not "very small"

in which case its more about L/R time constants. I = Vcc/R, so when you switch the solenoid off, the voltage across the internal inductance rises to Vcc (because of I*R) + Vclamp.

So the difference between 42+5 and 42+0.7 is bugger all, and the difference in decay time is, likewise, bugger all - well not bugger all, but certainly not 7x.


Cheers Terry

Reply to
Terry Given

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